It seems that the matrix $K\times K$ matrix $\boldsymbol M$ with components \begin{equation} M_{nj}=(n+1)^j \end{equation} is invertible for any dimension $K>0$.
I don't konw how to show this, but it seems natural. My first approach would be to show that the determinant is nonvanishing, i.e. \begin{equation} \det M=\sum_{\rho\in S_K}\text{sgn}({\rho})\prod^K_{n=1}(n+1)^{\rho(n)} \end{equation} or its transpose: \begin{equation} \det M^T=\sum_{\rho\in S_K}\text{sgn}({\rho})\prod^K_{n=1}(\rho(n)+1)^{n} \end{equation} but I don't see how to go from there. But I think it is already interesting in its own right.
The motivation stems from the following question: Is it true that for any sequence of real numbers $\zeta_1,\zeta_2,\dots, \zeta_K$ I can find a sequence of real numbers $\dots,s_{-3},s_{-2},s_{-1},s_{0},s_{1}s_{2},\dots$ such that $\zeta_k=\sum_{n\in \mathbb Z} s_{n}n^k$ for all $k$. If the above matrix is invertible, then this is the case since $\vec \zeta=\boldsymbol M^T \vec s $ has a solution (here we just need $s_1,\dots, s_K$, so the other entries could be chosen as 0 to get the infinite sequence).
However this is just the motivation and the initial question is more interesting.
$$ |M_{nj}|= \begin{vmatrix} 2^1 & 2^2 & 2^3 &...&2^n \\ 3^1 & 3^2 & 3^3 &...&3^n \\ ...\\ (n+1)^1 & (n+1)^2 & (n+1)^3 &...&(n+1)^n \\ \end{vmatrix}= $$
$$=2\cdot3\cdot4...\cdot(n+1) \begin{vmatrix} 1 & 2^1 & 2^2 &...&2^{n-1} \\ 1 & 3^1 & 3^3 &...&3^{n-1} \\ ...\\ 1 & (n+1)^1 & (n+1)^2 &...&(n+1)^{n-1} \\ \end{vmatrix}$$
So the last one is a Vandermonde Matrix wich is given by:
$$(n+1)!\cdot \prod_{1\le i\le j}(M_{i2}-M_{j2}) \ne0$$
once $M_{i2}\ne M_{j2}$.